A 75 kg man on a 39kg cart traveling at a speed of 2.3 m/s. He jumps off with zero horizontal speed realative to the ground. What is the resulting change in the speed of the cart?
I tried using m1v1=m2v2 but that didn't work out since I was expose to get 4.4 m/s as the answer. I can't figure out what i did wrong will you please show me how to solve this problem?
How much work is done by him?
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the man jumps off the cart is equal to the total momentum after he jumps off.
Before the man jumps off, the total momentum is given by:
(mass of man * velocity of man) + (mass of cart * velocity of cart)
After the man jumps off, the total momentum is given by:
(mass of cart * velocity of cart)
Since the man jumps off with zero horizontal velocity relative to the ground, we can consider his momentum contribution as zero.
Therefore, we can equate the two momenta:
(mass of man * velocity of man) + (mass of cart * velocity of cart) = (mass of cart * final velocity of cart)
Now, let's calculate the final velocity of the cart.
(mass of man * velocity of man) + (mass of cart * velocity of cart) = (mass of cart * final velocity of cart)
[75 kg * (0 m/s)] + [39 kg * (2.3 m/s)] = (39 kg * final velocity of cart)
0 + (39 kg * 2.3 m/s) = (39 kg * final velocity of cart)
89.7 kg·m/s = (39 kg * final velocity of cart)
Divide both sides of the equation by 39 kg:
89.7 kg·m/s / 39 kg = final velocity of cart
2.3 m/s = final velocity of cart
Therefore, the resulting change in the speed of the cart is 2.3 m/s.